Precalculus: Mathematics for Calculus, 7th Edition

by
Stewart, James; Redlin, Lothar; Watson, Saleem

Published by
Brooks Cole

ISBN 10:
1305071751

ISBN 13:
978-1-30507-175-9

Chapter 1 - Section 1.4 - Rational Expressions - 1.4 Exercises: 57

Answer

$\frac{-5}{(x-3)(x+1)(x+2)}$

Work Step by Step

Factor $x^{2}+3x+2$: $(x+1)(x+2)$
Factor $x^{2}-2x-3$: $(x+1)(x-3)$
Therefore, the two fractions become: $\frac{1}{(x+1)(x+2)}-\frac{1}{(x+1)(x-3)}$
Find the LCD for the two fractions. This is: $(x+2)(x+1)(x-3)$
Adjust the fractions based on the LCD:
$=\frac{x-3}{(x-3)(x+1)(x+2)}-\frac{(x+2)}{(x+2)(x+1)(x-3)}$
Since the denominators are equal, combine the fractions:
$=\frac{x-3-(x+2)}{(x-3)(x+1)(x+2)}$
Expand the numerator:
$=\frac{x-3-x-2}{(x-3)(x+1)(x+2)}$
Collect like terms:
$=\frac{-5}{(x-3)(x+1)(x+2)}$